Momentum and radiation pressure

Momentum rides with energy. For an electromagnetic wave, the momentum $p$ carried by an amount of energy $U$ is fixed by the wave's own speed, not by any mass:

where: $p$ is the momentum (kg·m/s) carried by the radiation, $U$ is the electromagnetic energy (J) in that same parcel of the wave, and $c=3\times10^8\ \text{m/s}$ is the speed of light. This single relation is the EM-wave analogue of the field relation $E=cB$ you already met for the fields themselves — it links two different wave properties through $c$, and everything below follows from it.

Absorption transfers $p$ once; reflection transfers it twice. If a surface absorbs the wave completely, the full incident momentum is handed over and stops there. If the surface reflects the wave straight back, the wave's momentum reverses direction, so the change in the wave's momentum — and hence the momentum given to the surface, by Newton's third law — is twice as large for the same incident energy. This is exactly why a rebounding (elastic) collision in mechanics delivers more momentum to a wall than a sticking (inelastic) one for the same incoming ball.

From momentum to pressure. Writing the incident power as $P=U/\Delta t$, and for a beam of intensity $I$ over area $A$ as $P=IA$, the force on an absorber is $F=\Delta p/\Delta t = P/c$, giving radiation pressure

where: $I$ is the intensity of the beam (W/m²) and $P_{\text{rad}}$ is the pressure (Pa) it produces. A surface that partially reflects (reflectance $R$, so a fraction $1-R$ is absorbed) interpolates between the two: $P_{\text{rad}}=\dfrac{I}{c}(1+R)$, correctly reducing to $I/c$ at $R=0$ and $2I/c$ at $R=1$. This course keeps to the two clean limits; the general $R$ formula is given only as context, not as something you need to derive. Because $c$ is so large, radiation pressure from ordinary light sources is tiny — micro-pascals for sunlight, forces of only micronewtons even from a strong laboratory laser. That smallness is why the effect needs careful apparatus (torsion balances) or astronomical scales (comet tails, dust pushed by starlight) to become visible, even though it is always present.

A laser beam of intensity $I=2\times10^{6}\ \text{W/m}^2$ is normally incident on a perfectly absorbing black surface of area $A=4\times10^{-4}\ \text{m}^2$ (4 cm²). Find (a) the radiation pressure and force on the black surface, (b) the same quantities if the surface is instead a perfect mirror, and (c) the extra force gained by switching from absorber to mirror.

Step 1. Absorber pressure. Using $P_{\text{rad}}=I/c$:

Step 2. Absorber force. Multiply pressure by area:

Already this confirms that laboratory radiation-pressure forces are micronewton-scale — far below hand sensitivity.

Step 3. Mirror pressure and force. Both quantities double, from $P_{\text{rad}}=2I/c$:

Step 4. Extra force from switching. The gain is simply the absorber's own force, since the mirror value is exactly double: $\Delta F = 5.3\times10^{-6} - 2.7\times10^{-6} = 2.7\times10^{-6}\ \text{N}$, i.e. reflecting recovers one extra "unit" of force equal to what absorption alone produced. The mirror feels $5.3\ \mu\text{N}$, twice the absorber's $2.7\ \mu\text{N}$, an extra $2.7\ \mu\text{N}$ from the reflected half of the momentum.

Competitive stretch

A 100 W isotropic bulb illuminates a small perfectly absorbing disc of area $1\ \text{cm}^2$ placed $2\ \text{m}$ away. Intensity spreads over a sphere of area $4\pi r^2$: $I = \dfrac{100}{4\pi(2)^2} \approx 1.99\ \text{W/m}^2$. Radiation pressure is $P_{\text{rad}}=I/c \approx 6.6\times10^{-9}\ \text{Pa}$, and the force on the $1\times10^{-4}\ \text{m}^2$ disc is $F = P_{\text{rad}}A \approx 6.6\times10^{-13}\ \text{N}$ — an almost immeasurably small push from an ordinary bulb, which is why radiation-pressure demonstrations favour intense collimated lasers or astronomical scales instead.

• Using $I/c$ automatically — the factor of 2 for a mirror is the single most-forgotten step; always decide absorber or reflector before writing the formula.
• Mixing up pressure and force — pressure ($I/c$ or $2I/c$) is per unit area; you must multiply by the actual illuminated area $A$ to get the force in newtons.
• Treating $p=U/c$ like ordinary $p=mv$ — EM radiation is massless, and $p=U/c$ is the correct relativistic substitute ($E=pc$), not a special case of mechanical momentum.
• Forgetting that only intensity striking the surface matters — at non-normal incidence only the component of momentum along the surface normal contributes to pressure; this course only requires normal incidence, so treat that as a caution, not something to derive.

Ask the tutor: if you would like to see how a partially reflecting surface's pressure $\dfrac{I}{c}(1+R)$ is derived, or how this momentum result connects to the photon picture of light.